
#1
Oct1107, 09:37 PM

P: 23

How do you find the Kinetic Energy of a person spinning on a chair with 2 equal weights on each arm?
A specific problem asks what the change in KE is if he moves his originally extended arms inward. I tried doing (1/2)(I)(wf)^2  (1/2)(I)(wi)^2 where wi is initial angular velocity and wf is final angular velocity, but its wrong. Any help? 



#2
Oct1107, 09:47 PM

HW Helper
P: 4,125

What numbers/variables do they give?




#3
Oct1107, 09:55 PM

P: 23

Ahh, sorry, I suppose I should have included this information in the main thread. My bad.
Well I is the same for initial and final, which is 8. Therefore, Initial I = 8 Final I = 8 Initial w = .7 Final w = 1.15 So I did: (1/2)(I)(wf)^2  (1/2)(I)(wi)^2 = (1/2)(8)(1.15)^2  (1/2)(8)(.7)^2 = 5.31118  1.96 = 3.35118116 j ...but apparently it's wrong. 



#4
Oct1107, 10:10 PM

HW Helper
P: 4,125

Finding Kinetic Energy of Person Spinning on Chair w/ Weights
I can't be the same initial and final... if he brings his arms in the moment of inertia changes. post the question exacty as it is stated...




#5
Oct1107, 10:28 PM

P: 23

Alright. But just fyi, I know it cant be the same I because the r changes, but the problem says to just assume its the same. I used the same I for initial and final to get the wf, and it was correct, so i know I is same initial and final. Anyways, heres the problem:
A student sits on a rotating stool holding two 3 kg objects. When his arms are extended horizontally, the objects are 1 m from the axis of rotation, and he rotates with angular speed of 0.7 rad/sec. The moment of inertia of the student plus the stool is 8 kgm^2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.29 m from the rotation axis. a) Calculate the final angular speed of the student. Answer in units of rad/s. (The answer to this part is 1.15232 rad/s) b) Calculate the change in kinetic energy of the system. Answer in units of J. 


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